D. 72.4 g
To solve this problem, we first need to determine the limiting reactant by calculating how much AC3 can be produced from 17.6 g of AB.
1. Calculate the molar mass of AB and AC3:
AB = 40 g/mol
AC3 = 56 g/mol
2. Convert the mass of AB to moles:
17.6 g AB x (1 mol AB/40 g AB) = 0.44 mol AB
3. Use the mole ratio from the balanced equation to find the moles of AC3 produced:
1 mol AB produces 2 mol AC3
0.44 mol AB x (3 mol AC3/1 mol AB) = 1.32 mol AC3
4. Convert moles of AC3 to grams:
1.32 mol AC3 x (56 g AC3/1 mol AC3) = 73.92 g AC3
Therefore, the mass of AC3 produced from 17.6 g of AB is 73.92 g.
The closest answer choice is 72.4 g, so the correct answer is D. 72.4 g.
Using the same hypothetical, unbalanced equation from problem 8, which of the following mass in grams of AC3 is produced when 17.6 g of AB reacts with unlimited C2?
A. 36.2 g
B. 54.4 g
C. 61.2 g
1 answer